Originally Posted by

**arbolis** I've followed Chris' tutorial, but I didn't get lucky at the end.

I must solve $\displaystyle y''+2y'=xe^{-2x}$. Also, $\displaystyle y(0)=a$ and $\displaystyle y'(0)=b$.

So I consider the homogeneous equation $\displaystyle y''+2y'=0$. I wrote the characteristic polynomial and found its roots to be $\displaystyle \pm i \sqrt 2$. Thus a solution to the homogeneous DE is $\displaystyle c_1 \cos (x\sqrt 2)+c_2 \sin (x \sqrt 2)$. I'm not sure if I can solve for $\displaystyle c_1$ and $\displaystyle c_2$ yet, with the initial conditions. It would be the solution the homogeneous DE, but are these constants the same in the case of the non homogeneous case? I guess yes...

Anyway, I continued writing $\displaystyle (D^2+2D)(y)=xe^{-2x}$. The annihilator is $\displaystyle (D+2)$.

Hence $\displaystyle (D+2)(D^2+2D)(y)=0$. So $\displaystyle (r+2)(r^2+2r)=0 \Rightarrow $$\displaystyle r=0$ or $\displaystyle r=-2$ (this one is a double root).

Thus a general solution to the original DE is $\displaystyle y=c_1 \cos (x\sqrt 2)+c_2 \sin (x \sqrt 2)+c_3e^{-2x}+c_4$.

I tried to solve for $\displaystyle c_3$ and $\displaystyle c_4$ this way:

$\displaystyle y_p=c_3e^{-2x}$.

$\displaystyle y_p'=-2c_3e^{-2x}$.

$\displaystyle y_p''=4c_3e^{-2x}$.

Plugging these values into the original DE, I reach that $\displaystyle xe^{-2x}=0, c_3 $ cancels out... Obviously I made something wrong. I don't see what.